Condition for Two Given Numbers to Be Roots

We start with the quadratic equation $$81x^2 + kx + 256 = 0$$ whose two real roots we denote by $$\alpha$$ and $$\beta$$. According to the statement, one root is the cube of the other. We can therefore set

$$\beta = \alpha^3.$$

For any quadratic equation of the form $$ax^2 + bx + c = 0$$ we have the standard Vieta relations:

• Sum of the roots: $$\alpha + \beta = -\dfrac{b}{a}.$$
• Product of the roots: $$\alpha\beta = \dfrac{c}{a}.$$

In our particular equation $$a = 81,\; b = k,\; c = 256.$$ So the relations become

$$\alpha + \beta = -\dfrac{k}{81} \quad\text{and}\quad \alpha\beta = \dfrac{256}{81}.$$

Substituting $$\beta = \alpha^3$$ into the product relation gives

$$\alpha \cdot \alpha^3 = \dfrac{256}{81},$$

so

$$\alpha^4 = \dfrac{256}{81}.$$

We recognise that $$256 = 4^4$$ and $$81 = 3^4,$$ hence

$$\alpha^4 = \left(\dfrac{4}{3}\right)^4.$$

Taking the real fourth root on both sides, we obtain the two possible values

$$\alpha = \dfrac{4}{3} \quad\text{or}\quad \alpha = -\dfrac{4}{3}.$$

Now we use the sum relation $$\alpha + \beta = -\dfrac{k}{81}.$$ Remembering that $$\beta = \alpha^3,$$ we handle the two cases separately.

Case 1: $$\alpha = \dfrac{4}{3}.$$ Then $$\beta = \left(\dfrac{4}{3}\right)^3 = \dfrac{64}{27}.$$ Hence

$$\alpha + \beta = \dfrac{4}{3} + \dfrac{64}{27} = \dfrac{36}{27} + \dfrac{64}{27} = \dfrac{100}{27}.$$

Putting this into $$\alpha + \beta = -\dfrac{k}{81}$$ we get

$$\dfrac{100}{27} = -\dfrac{k}{81}.$$

Cross-multiplying gives

$$100 \times 81 = -27k,$$ so

$$k = -\dfrac{100 \times 81}{27} = -100 \times 3 = -300.$$

Case 2: $$\alpha = -\dfrac{4}{3}.$$ Then $$\beta = \left(-\dfrac{4}{3}\right)^3 = -\dfrac{64}{27}.$$ Thus

$$\alpha + \beta = -\dfrac{4}{3} - \dfrac{64}{27} = -\dfrac{36}{27} - \dfrac{64}{27} = -\dfrac{100}{27}.$$

Substituting into $$\alpha + \beta = -\dfrac{k}{81}$$ gives

$$-\dfrac{100}{27} = -\dfrac{k}{81}.$$

Canceling the negative signs we get

$$\dfrac{100}{27} = \dfrac{k}{81},$$

which leads to

$$k = \dfrac{100 \times 81}{27} = 100 \times 3 = 300.$$

Among the two values $$k = -300$$ and $$k = 300,$$ only $$-300$$ appears in the given list of options.

Hence, the correct answer is Option D.

Let $$\alpha$$ and $$\beta$$ be the roots of the equation.

$$\Rightarrow$$ $$\alpha + \beta = -4$$ and $$\alpha\beta = -n$$

The roots of the equation should be integral.

We want pairs such that the sum of the roots is $$-4$$ and the product of the roots is $$-n$$, where $$n$$ lies in $$[20, 100]$$. Hence, the following possible pairs exist. We can enumerate the cases and find out the number of distinct values of $$n$$

$$(3, -7) \Rightarrow n = 21$$

$$(4, -8) \Rightarrow n = 32$$

$$(5, -9) \Rightarrow n = 45$$

$$(6, -10) \Rightarrow n = 60$$

$$(7, -11) \Rightarrow n = 77$$

$$(8, -12) \Rightarrow n = 96$$

This gives 6 possible values of $$n$$

Thus, option C is the correct choice.

Let the four numbers be in the same order as stated: $$\alpha ,\beta ,\gamma ,\delta$$ form a geometric progression (G.P.) with common ratio $$r$$.

Hence
$$\beta = \alpha r,\qquad \gamma = \alpha r^{2},\qquad \delta = \alpha r^{3}$$

The roots of the first equation $$x^{2}-x+p=0$$ satisfy
$$\alpha + \beta = 1,\qquad \alpha \beta = p$$

The roots of the second equation $$x^{2}-4x+q=0$$ satisfy
$$\gamma + \delta = 4,\qquad \gamma \delta = q$$

Using the G.P. relations, write the two sums in terms of $$\alpha$$ and $$r$$:
$$\alpha + \beta = \alpha + \alpha r = \alpha(1+r) = 1 \qquad -(1)$$
$$\gamma + \delta = \alpha r^{2} + \alpha r^{3} = \alpha r^{2}(1+r) = 4 \qquad -(2)$$

Divide $$(2)$$ by $$(1)$$:
$$\frac{\alpha r^{2}(1+r)}{\alpha(1+r)} = r^{2} = 4 \;\Longrightarrow\; r = \pm 2$$

Case 1: $$r = 2$$

From $$(1)$$, $$\alpha = \dfrac{1}{1+2} = \dfrac{1}{3}$$.
Then $$p = \alpha^{2} r = \left(\dfrac{1}{3}\right)^{2}\!\cdot 2 = \dfrac{2}{9} \notin \mathbb{Z}$$. This violates the condition $$p \in \mathbb{Z}$$, so $$r = 2$$ is rejected.

Case 2: $$r = -2$$

From $$(1)$$, $$\alpha = \dfrac{1}{1-2} = -1$$.

Now compute the required products:
$$p = \alpha^{2} r = (-1)^{2}(-2) = -2$$
$$q = \gamma \delta = (\alpha r^{2})(\alpha r^{3}) = \alpha^{2} r^{5} = (-1)^{2}(-2)^{5} = -32$$

Both $$p$$ and $$q$$ are integers, so this case is admissible.

Finally,
$$p + q = -2 + (-32) = -34 \;\Longrightarrow\; |p+q| = 34$$

Therefore, the required value is $$34$$.

Option C which is: $$34$$